Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does any $u \in \mathbb{C}$ exist such that:

$$\frac{u}{\sqrt{-u^2}}=1$$

If yes, give an example please.

UPDATE:

OK, I thought a little about that myself and I think it goes like this ($m,n\in\mathbb{N}$ and $r,\phi\in\mathbb{R}$):

$$\frac{u}{\sqrt{-u^2}}=\frac{|r|e^{i(\phi+2\pi n)}}{\sqrt{e^{i(\pi+2\pi m)}(|r|e^{i(\phi+2\pi n)})^2}}=\frac{e^{i(\phi+2\pi n)}}{e^{i\frac{\pi+2\pi m}{2}}e^{i (\phi+2\pi n)}}=e^{-i\pi(\frac{1}{2}+ m)}$$

Since $(\frac{1}{2}+m)$ can never be an even integer, the above equation can never hold.

share|improve this question
add comment

3 Answers

No such solution can exist. Squaring both sides gives $$\frac{u^2}{-u^2}=1$$ which can only hold true if $1 = -1$.

share|improve this answer
    
Could you elaborate a little more on why one apparently can generally assume that $\frac{u}{\sqrt{-u^2}}=\sqrt{\frac{u^2}{-u^2}}$? Your solution is only true if that holds. –  bollty Jan 24 '13 at 10:52
    
@bollty We have $a = b \implies a^2 = b^2$, so $(\frac{u}{\sqrt{-u^2}})^2 = 1^2$ is implied by the statement, and this gives my equation without taking the square root of anything. –  Sam DeHority Jan 24 '13 at 10:58
    
Same story, please elaborate on why one generally can assume $(\frac{u}{\sqrt{-u^2}})^2=\frac{(u)^2}{(\sqrt{-u^2})^2}$. I think with complex numbers this step is not trivial. –  bollty Jan 24 '13 at 11:07
    
$(\frac{z}{w})^2 = (\frac{z \bar{w}}{w \bar{w}})^2 = \frac{z^2 \bar{w}^2}{w^2 \bar{w}^2} = \frac{z^2}{w^2}$ for complex $z,w$ –  Sam DeHority Jan 24 '13 at 11:34
    
That would again ask for the verification of the $\left(\frac{z^2}{w^2}\right)\cdot\left(\frac{{\bar w}^2}{{\bar w}^2}\right)=\frac{(z^2 {\bar w}^2)}{(w^2{\bar w}^2)}$, which works well in the exponential representation. –  bollty Jan 24 '13 at 12:14
show 1 more comment

Proceed like this:

$$u = \sqrt{-u^2}$$

Squaring both sides:

$$u ^ 2 = - u ^2 $$

or $$1 = -1$$

which is contradiction. Hence, not possible.

share|improve this answer
1  
Please note, your altering of the shape has the trivial solution $u=0$ which is not true for the initial equation. –  bollty Jan 24 '13 at 10:50
1  
Switching sides/cancellation/division by u is applicable only with $u \neq 0$ –  hjpotter92 Jan 24 '13 at 10:52
add comment

$$\frac{u}{\sqrt{-u^2}}\cdot\frac{u}{\sqrt{-u^2}}=1.1 \implies\frac{u^2}{-u^2}=1 \implies -1=1$$ Your equation only give the solution $1=-1$. So It is not a valid equation whether $u \in \mathbb{C}$ or not.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.